Question
A proton and an α-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λp and λa related to each other?
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Answer
Key concept: Hence de-Broglie wavelength: $\lambda=\frac{\text{h}}{\text{p}}=\frac{\text{h}}{\sqrt{2\text{mE}}}=\frac{\text{h}}{\sqrt{2\text{mqV}}}$ In this problrm since both proton and $\alpha-\text{particle}$ are accelerated through same potential difference, We know that, $\lambda=\frac{\text{h}}{\sqrt{2\text{mqv}}}$ $\therefore\ \lambda\propto\frac{1}{\sqrt{\text{mq}}}$ $\frac{\lambda_\text{p}}{\lambda_\alpha}=\frac{\sqrt{\text{m}_\alpha\text{q}_\alpha}}{\text{m}_\text{p}\text{q}_\text{p}}=\frac{\sqrt{4\text{m}_\text{p}\times2\text{e}}}{\sqrt{\text{m}_\text{p}\times\text{e}}}=\sqrt{8}$ $\therefore\ \lambda_\text{p}=\sqrt{8}\lambda_\alpha$ i.e., wavelength of proton is times wavelength of $\alpha-\text{particle}$.
Important point:
De-Broglie wavelength associated with the charged particles: The energy of a charged particle acceletated through potential difference V is $\text{E}=\frac{1}{2}\text{mv}^2=\text{qV}$ Hence de-Broglie wavelength $\lambda=\frac{\text{h}}{\text{p}}=\frac{\text{h}}{\sqrt{2\text{mE}}}=\frac{\text{h}}{\sqrt{2\text{mqV}}}$ $\lambda_\text{Electron}=\frac{12.27}{\sqrt{\text{V}}}\mathring{\text{A}},\lambda_\text{Proton}=\frac{0.286}{\sqrt{\text{V}}}\mathring{\text{A}}$ $\lambda_\text{Deutron}=\frac{0.200}{\sqrt{\text{V}}}\mathring{\text{A}},\lambda_{\alpha-\text{particle}}=\frac{0.101}{\sqrt{\text{V}}}\mathring{\text{A}}$Need a full question paper?
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